r^2+12=50

Solution for r^2+12=50 equation:

r^2+12=50

We move all terms to the left:

r^2+12-(50)=0

We add all the numbers together, and all the variables

r^2-38=0

a = 1; b = 0; c = -38;
Δ = b2-4ac
Δ = 02-4·1·(-38)
Δ = 152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{38}}{2*1}=\frac{0-2\sqrt{38}}{2}
=-\frac{2\sqrt{38}}{2}
=-\sqrt{38}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{38}}{2*1}=\frac{0+2\sqrt{38}}{2}
=\frac{2\sqrt{38}}{2}
=\sqrt{38}
$